Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = (int)1e5 + 3;
const int MAXLOG = 17;
int lca[MAXLOG][MAXN], N, S, Q, height[MAXN], par[MAXN];
vector<int> soldier[MAXLOG][MAXN], adj[MAXN];
void dfs(int curr, int h, int p) {
  par[curr] = p;
  height[curr] = h;
  for (int i = 0; i < adj[curr].size(); i++) {
    int v = adj[curr][i];
    if (v == p) continue;
    dfs(v, h + 1, curr);
  }
}
void mrg(vector<int>& a, vector<int>& b, vector<int>& res, int k) {
  res.resize(a.size() + b.size());
  merge(a.begin(), a.end(), b.begin(), b.end(), res.begin());
  while (res.size() > k) res.pop_back();
}
void prec() {
  for (int i = MAXLOG - 1; i >= 0; i--) {
    for (int j = 1; j <= N; j++) lca[i][j] = (i == 0 ? par[j] : -1);
  }
  for (int i = 1; i < MAXLOG; i++) {
    for (int j = 1; j <= N; j++) {
      if (lca[i - 1][j] != -1) {
        lca[i][j] = lca[i - 1][lca[i - 1][j]];
        mrg(soldier[i - 1][j], soldier[i - 1][lca[i - 1][j]], soldier[i][j],
            10);
      }
    }
  }
}
void printResult(int k, vector<int>& res) {
  printf("%d", min(k, (int)res.size()));
  for (int i = 0; i < min(k, (int)res.size()); i++) {
    printf(" %d", res[i]);
  }
  printf("\n");
}
void solve(int u, int v, int k) {
  if (height[u] < height[v]) swap(u, v);
  vector<int> res, tmp;
  for (int i = MAXLOG - 1; i >= 0; i--) {
    if (height[u] - (1 << i) >= height[v]) {
      mrg(tmp, soldier[i][u], res, k);
      tmp = res;
      u = lca[i][u];
    }
  }
  if (u == v) {
    mrg(tmp, soldier[0][u], res, k);
    printResult(k, res);
    return;
  }
  for (int i = MAXLOG - 1; i >= 0; i--) {
    if (lca[i][u] != -1 && lca[i][u] != lca[i][v]) {
      mrg(tmp, soldier[i][u], res, k);
      tmp = res;
      mrg(tmp, soldier[i][v], res, k);
      tmp = res;
      u = lca[i][u];
      v = lca[i][v];
    }
  }
  mrg(tmp, soldier[0][u], res, k);
  tmp = res;
  mrg(tmp, soldier[0][v], res, k);
  tmp = res;
  mrg(tmp, soldier[0][par[u]], res, k);
  printResult(k, res);
}
int main() {
  scanf("%d%d%d", &N, &S, &Q);
  for (int i = 0, a, b; i < N - 1; i++) {
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, 0, -1);
  for (int i = 1, x; i <= S; i++) {
    scanf("%d", &x);
    soldier[0][x].push_back(i);
  }
  for (int i = 1; i <= N; i++) {
    if (!soldier[0][i].empty())
      sort(soldier[0][i].begin(), soldier[0][i].end());
    while (soldier[0][i].size() > 10) soldier[0][i].pop_back();
  }
  prec();
  for (int q = 0, u, v, k; q < Q; q++) {
    scanf("%d%d%d", &u, &v, &k);
    solve(u, v, k);
  }
  return 0;
}